Tag: physics

Questions Related to physics

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1 psi is equivalent to $-$ atmospheric pressure.

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $34.046\times 10^{-3}$

  2. $64.046\times 10^{-6}$

  3. $24.046\times 10^{-3}$

  4. $68.046\times 10^{-3}$

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Correct Option: D
Explanation:

Psi stands for pound force per square inch. 

Normal atmospheric pressure is 14.7psi.
         $14.7psi=1$ atmospheric pressure

         $1psi=68.027\times10^{-3}atm$

$1\ Bar$ of pressure is equal to:

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $1\ kPa$

  2. $10\ kPa$

  3. $100\ kPa$

  4. $1000\ kPa$

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Correct Option: C
Explanation:

Atmospheric pressure $=1atm=1.01\times 10^5Pa=1.01bar$

$\implies 1bar=10^5Pa=100kPa$

In free space, Fortin's barometer will read:

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $0\ mm$ of Hg

  2. $76\ cm$ of Hg

  3. $1\ cm$ of Hg

  4. None of these

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Correct Option: A
Explanation:

Fortin's barometer is a modified form of Torricelli's barometer which is used to measure atmospheric pressure. In free space there is no matter (no atmosphere) therefore no atmospheric  pressure hence barometer will read 0mm of Hg .

The pressure of a gas filled in the bulb of constant volume gas thermometer at $0^0C$ and $100^0C$ are 28.6 cm and 36.6 cm of mercury respectively. The temperature of bulb at which pressure will be 35.00 cm of mercury will be:

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $80^0C$

  2. $70^0C$

  3. $55^0C$

  4. $40^0C$

Show answer
Correct Option: A
Explanation:

from$T=\dfrac{p-p _0}{p _{100}-p _0}\times 100^0=\dfrac{35-28.6}{36.6-28.6}\times 100^0=80^0$

The pressure of a gas filled in the bulb of a constant volume gas thermometer at $0^o$C and $100^o$C are $28.6cm$ and $36.6cm$ of mercury respectively. The temperature of bulb at which pressure will be $35.0cm$ of mercury will be 

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $80^o$C

  2. $70^o$C

  3. $55^o$C

  4. $40^o$C

Show answer
Correct Option: A
Explanation:
In a constant volume gas thermometer, the pressure of the gas varies in proportion to the temperature of the gas. The temperature varies linearly wrt to the pressure and vice versa.

 So the formula is:   $ T = \dfrac{(P - P _{0})}{(P _{100} - P _{0})}\times 100^\circ C$

     Where $T$ is the temperature at pressure $P$.
                $P _0$ is pressure at $T _0 = 0^\circ C$ 
             and $P _{100}$ is pressure at $T _{100} = 100⁰C$
                 
   $T = \dfrac{(35.0 - 28.6)}{(36.6 - 28.6)}  \times 100 ^\circ C$  
     
$  = 80^\circ C$

The height of mercury in a barometer is $760\ mm$, then the atmosphere pressure is $[g=9.8\ {m/s}^{2}]$:

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $0.101\times {10}^{5}Pa$

  2. $10.1\times {10}^{5}Pa$

  3. $1. 01\times {10}^{5}Pa$

  4. $1.01\times {10}^{4}Pa$

Show answer
Correct Option: C
Explanation:

$P=\rho gh=13600\times 9.8\times 0.760=1.01\times 10^5Pa$


Two communicating cylindrical vessel contain mercury. The diameter of one vessel is n times larger than the diameter of the other.A column of water of height h is poured into the vessel.The mercury level will rise in the right - hand vessel (s=relative density of mercury and p = density of water ) by 

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $ \frac { n^ h }{ (n\quad +\quad 1)^ 2s } $

  2. $ \frac { h }{ (n^ 2\quad +\quad 1)s } $

  3. $ \frac { h }{ (n\quad +\quad 1)^ 2s } $

  4. $ \frac { h }{ (n^{ 2 }s) } $

Show answer
Correct Option: B
Explanation:

It is given that the diameter of one vessel is n times larger than the diameter of other. So,

$ \pi {{r}^{2}}{{h} _{1}}=\pi {{(nr)}^{2}}{{h} _{2}} $

$ {{h} _{1}}={{n}^{2}}{{h} _{2}} $

Since, pressure at point A is equal to pressure at B

$ \rho gh=({{h} _{1}}+{{h} _{2}})\rho 'g $

$ \because s=\dfrac{\rho '}{\rho } $

$ \rho gh=({{n}^{2}}{{h} _{2}}+{{h} _{2}})s\rho g $

$ {{h} _{2}}=\dfrac{h}{({{n}^{2}}+1)s} $

A thin tube sealed at both ends, is $100\ cm$ long. It lies horizontally, the middle $0.1\ m$ containing mercury and the two ends containing air at standard atmospheric pressure. If the tube is turned to a vertical position, by what amount will the mercury be displaced ?

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $1.84\ cm$

  2. $8.45\ cm$

  3. $2.95\ cm$

  4. $5\ cm$

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Correct Option: C

A barometer tube reads $76$ $cm$ of mercury. If the tube is gradually inclined at an angle of $60 ^ { \circ }$ with vertical, keeping the open end immersed in the mercury reservoir, the length of the mercury column will be  

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. $152 \mathrm { cm }$

  2. $76 \mathrm { cm }$

  3. $\dfrac{76cm}{sin30^o}$

  4. $38 \mathrm { cm }$

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Correct Option: B

2.56g of sulphur (colloidal sol) in 100 ml solution shows Osmotic pressure of 2.463 atm at ${ 27 }^{ 0 }C$. How many sulphur atoms are associated in colloidal sol ? [Solution constant = 0.0821 atm.${ mol }^{ -1 }{ k }^{ -1 }$]

physics fluid pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

  1. 2

  2. 4

  3. 5

  4. 8

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Correct Option: C